How to find a number of solutions for $x_1+x_2+ldots+x_n=r$ where $r$ is divisible by 3 and for a given $l$...
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How to find a number of solutions for $x_1+x_2+ldots+x_n=r$ where $r$ is divisible by 3 and for a given $l$ and $r$ such that $lleq x_i leq r$?
combinatorics combinations
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add a comment |
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How to find a number of solutions for $x_1+x_2+ldots+x_n=r$ where $r$ is divisible by 3 and for a given $l$ and $r$ such that $lleq x_i leq r$?
combinatorics combinations
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1
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what is this have to do with Taylor expansion?
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– pointguard0
Jan 20 at 14:02
2
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Letting $y_i=x_i-l$ this becomes a routine Stars and Bars problem. $r$ can be whatever you want...divisibility by $3$ doesn't change the computation.
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– lulu
Jan 20 at 14:10
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Or maybe using generating functions, similar to this.
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– rtybase
Jan 21 at 12:52
add a comment |
$begingroup$
How to find a number of solutions for $x_1+x_2+ldots+x_n=r$ where $r$ is divisible by 3 and for a given $l$ and $r$ such that $lleq x_i leq r$?
combinatorics combinations
$endgroup$
How to find a number of solutions for $x_1+x_2+ldots+x_n=r$ where $r$ is divisible by 3 and for a given $l$ and $r$ such that $lleq x_i leq r$?
combinatorics combinations
combinatorics combinations
edited Jan 21 at 11:17
N. F. Taussig
44.3k93357
44.3k93357
asked Jan 20 at 13:58
Sail AkhilSail Akhil
205
205
1
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what is this have to do with Taylor expansion?
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– pointguard0
Jan 20 at 14:02
2
$begingroup$
Letting $y_i=x_i-l$ this becomes a routine Stars and Bars problem. $r$ can be whatever you want...divisibility by $3$ doesn't change the computation.
$endgroup$
– lulu
Jan 20 at 14:10
$begingroup$
Or maybe using generating functions, similar to this.
$endgroup$
– rtybase
Jan 21 at 12:52
add a comment |
1
$begingroup$
what is this have to do with Taylor expansion?
$endgroup$
– pointguard0
Jan 20 at 14:02
2
$begingroup$
Letting $y_i=x_i-l$ this becomes a routine Stars and Bars problem. $r$ can be whatever you want...divisibility by $3$ doesn't change the computation.
$endgroup$
– lulu
Jan 20 at 14:10
$begingroup$
Or maybe using generating functions, similar to this.
$endgroup$
– rtybase
Jan 21 at 12:52
1
1
$begingroup$
what is this have to do with Taylor expansion?
$endgroup$
– pointguard0
Jan 20 at 14:02
$begingroup$
what is this have to do with Taylor expansion?
$endgroup$
– pointguard0
Jan 20 at 14:02
2
2
$begingroup$
Letting $y_i=x_i-l$ this becomes a routine Stars and Bars problem. $r$ can be whatever you want...divisibility by $3$ doesn't change the computation.
$endgroup$
– lulu
Jan 20 at 14:10
$begingroup$
Letting $y_i=x_i-l$ this becomes a routine Stars and Bars problem. $r$ can be whatever you want...divisibility by $3$ doesn't change the computation.
$endgroup$
– lulu
Jan 20 at 14:10
$begingroup$
Or maybe using generating functions, similar to this.
$endgroup$
– rtybase
Jan 21 at 12:52
$begingroup$
Or maybe using generating functions, similar to this.
$endgroup$
– rtybase
Jan 21 at 12:52
add a comment |
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1
$begingroup$
what is this have to do with Taylor expansion?
$endgroup$
– pointguard0
Jan 20 at 14:02
2
$begingroup$
Letting $y_i=x_i-l$ this becomes a routine Stars and Bars problem. $r$ can be whatever you want...divisibility by $3$ doesn't change the computation.
$endgroup$
– lulu
Jan 20 at 14:10
$begingroup$
Or maybe using generating functions, similar to this.
$endgroup$
– rtybase
Jan 21 at 12:52